# Purely unrectifiable sets#

Definition 24 (Purely unrectifiable set)

A $$\mathcal H^n$$-measurable set $$S\subset \mathbb R^m$$ is purely $$n$$-unrectifiable if, for all $$n$$-rectifiable $$E\subset \mathbb R^n$$, $$\mathcal H^n(S\cap E)=0$$.

Lemma 10

The four corner Cantor set $$K\subset \mathbb R^2$$ is purely 1-unrectifiable.*

Proof. Observe that the coordinate projections of $$K$$ have $$\mathcal L^1$$ measure zero (see Example 48). If there existed a rectifiable $$\gamma\subset \mathbb R^2$$ with $$\mathcal H^1(\gamma\cap K)>0$$, then $$\gamma\cap K$$ is a 1-rectifiable set of positive measure and hence, by Corollary 2, one of the coordinate projections must have positive measure, a contradiction.

For a second proof see Example 49.

Lemma 11

Let $$A\subset \mathbb R^m$$ be $$\mathcal H^m$$-measurable with $$\mathcal H^m(A)<\infty$$. There exists a decomposition $$A=E\cup S$$ with $$E$$ $$n$$-rectifiable and $$S$$ purely $$n$$-unrectifiable.*

Proof. Let

$t = \sup\{\mathcal H^n(E): E\subset A, n\text{-rectifiable}\}.$

Since $$\mathcal H^n(A)<\infty$$, $$t<\infty$$. Let $$E_i\subset A$$ be $$n$$-rectifiable with $$\mathcal H^n(E_i)\to t$$. Then $$E=\bigcup_{i\in \mathbb N}E_i$$ is $$n$$-rectifiable and is contained in $$A$$. Therefore

$t\geq \mathcal H^n(E)\geq \mathcal H^n(E_i)\to t$

and so $$\mathcal H^n(E)=t$$. Then $$S=A\setminus E$$ is purely $$n$$-unrectifiable. Indeed, if $$f\colon \mathbb R^n\to \mathbb R^m$$ is Lipschitz, $$E'=(A\setminus E)\cap f(\mathbb R^n)$$, is $$\mathcal H^m$$-measurable and so

$t\geq \mathcal H^n(E\cup E')=\mathcal H^n(E)+\mathcal H^n(E')= t +\mathcal H^n(E').$

We now state a very important theorem on the structure of purely unrectifiable sets. It requires the notion of a natural measure $$\gamma_{n,m}$$ on $$G(n,m)$$ that is invariant under the action of $$SO(m)$$. There are several ways to construct this measure. The simplest is to consider $$G(n,m)$$ as a (compact) metric space equipped with the metric

$d(V,W)=\|\pi_V-\pi_W\|.$

Then $$\gamma_{n,m}$$ is given by (a scalar multiple of) $$\mathcal H^{n(m-n)}$$. We will not discuss the specific details of this measure. When $$n=1$$, we may identify $$G(1,m)$$ with $$\mathbb S^{m-1}$$. In this case, $$\gamma_{1,m}$$ is simply $$\mathcal H^{m-1}$$.

Theorem 15 (Besicovitch–Federer projection theorem)

Let $$S\subset \mathbb R^m$$ be purely $$n$$-unrectifiable with $$\mathcal H^n(S)<\infty$$. Then, for $$\gamma_{n,m}$$-a.e. $$V\in G(n,m)$$,

$\mathcal L^n(\pi_V(S))=0.$

Conversely, if $$E\subset \mathbb R^m$$ is purely $$n$$-unrectifiable with $$\mathcal H^n(E)>0$$, for $$\gamma_{n,m}$$-a.e. $$V\in G(n,m)$$,

$\mathcal L^n(\pi_V(E))>0.$

Remark 6

The converse statement is given by Corollary 2.

We will prove the projection theorem for $$n=1$$ and $$m=2$$, which was proved by Besicovitch. First we prove some preliminary geometric properties of purely unrectifiable sets.

Lemma 12

Let $$E\subset \mathbb R^m$$, $$V\in G(m-n,m)$$ and $$0<s<1$$. Suppose that, for every $$x\in E$$,

$E\cap C(x,V,s)\cap B(x,r)=\emptyset.$

Then $$E$$ is $$n$$-rectifiable.

Proof. Since $$E$$ may be divided into countably many sets of diameter at most $$r$$, we may suppose $$\operatorname{diam}E\leq r$$. In this case, $$\pi_{V^\perp}$$ has Lipschitz inverse on $$E$$. Indeed, if $$x,y\in E$$ then $$y\not\in C(x,V,s)$$ and so

$\|\pi_{V^\perp}(x-y)\| = \operatorname{dist}(x-y,V)\geq s\|x-y\|.$

Therefore, $$E$$ is covered by a Lipschitz image of $$\mathbb R^n$$.

Lemma 13

Let $$S\subset \mathbb R^m$$ be purely $$n$$-unrectifiable, $$V\in G(m-n,m)$$, $$0<s<1$$ and $$0<\delta,\lambda<\infty$$. If

(24)#$\mathcal H^n(S\cap C(x,V,s)\cap B(x,r))\leq \lambda r^n s^n$

for every $$x\in S$$ and $$0<r<\delta$$ then

$\mathcal H^n(S \cap B(a,\delta/6))\leq 2\cdot 20^n \lambda \delta^n$

for every $$a\in S$$.

Remark 7

Note that this is certainly not true for a rectifiable set $$E$$; the first cone may be empty for every $$x\in E$$.

Proof. For a fixed $$a\in S$$, we may suppose $$S\subset B(a,\delta/6)$$. By Lemma 12, we may suppose that

$S\cap C(x,V,s/4)\neq\emptyset$

for every $$x\in S$$. For every $$x\in S$$ let

$h(x)=\sup \{|x-y|: y\in S \cap C(x,V,s/4)\},$

so that $$0<h(x)<\delta/3$$ for all $$x\in S$$. Pick $$x^*\in S\cap C(x,V,s/4)$$ with $$|x-x^*|\geq 3h(x)/4$$ and let $$C_x$$ be the cylinder

$C_x = \pi^{-1}_{V^\perp}(\pi_{V^\perp}(B(x,sh(x)/4))).$

We claim that

(25)#$C_x \cap S\subset C(x,V,s)\cap B(x,2h(x)) \cup C(x^*,V,s)\cap B(x^*,2h(x)).$

(Draw a picture!) Suppose $$z\in C_x \cap S$$ does not belong to the first set. Then

$\begin{split}s\|x^*-z\| &\leq \|\pi_{V^\perp}(x^*-z)|\\ &\leq \|\pi_{V^\perp}(x^*-x)\| + \|\pi_{V^\perp}(x-z)\|\\ &\leq s|x^*-x|/4 + sh(x)/4\\ &\leq sh(x)/2,\end{split}$

where the penultimate inequality follows because $$x^*\in C(x,V,s/4)$$ and $$z\in C_x$$. Therefore

$\begin{split}\|x-z\| &\geq \|x-x^*\| - \|x^*-z\|\\ &> 3h(x)/4 - h(x)/2\\ &\geq \|\pi_{V^\perp}(x-z)\|/s.\end{split}$

That is, $$z$$ belongs to the first set in (25).

By (25) and (24),

$\mathcal H^1(S\cap C_x) \leq 2\lambda (2h(x)s)^n.$

We apply Lemma 4 to the balls

$\pi_{V^\perp}(B(x,sh(x)/20))$

with $$x\in S$$. This gives countably many $$x_i\in S$$, for which these balls are disjoint, and

$S \subset \bigcup_{i\in\mathbb N}C_{x_i}.$

Therefore

$\begin{split}\mathcal H^n(S) &\leq \sum_{i\in\mathbb N} \mathcal H^n(C_{x_i})\\ &\leq 2\lambda 2^n \sum_{i\in\mathbb N} (sh(x_i))^n\\ &= 2 \lambda 2^n 20^n \sum_{i\in\mathbb N} \left(\frac{sh(x_i)}{20}\right)^n.\end{split}$

But, the $$\pi_{V^\perp}(B(x_i,sh(x_i)/20))$$ are disjoint subsets of $$B(\pi_{V^\perp}(a),\delta/2)\subset V^\perp$$ and so the final sum is bounded above by $$(\delta/2)^n$$.

Corollary 3

If $$S\subset \mathbb R^m$$ is purely $$n$$-unrectifiable with $$\mathcal H^n(S)<\infty$$ then for every $$V\in G(m-n,m)$$, every $$0<s<1$$ and $$\mathcal H^n$$-a.e. $$x\in S$$,

$\Theta^{*,n}(S\cap C(a,V,s),a) \geq 240^{-n-1} s^n.$

Proof. For a fixed $$V,s$$, this is immediate from the fact that $$\Theta^{*,n}(S,a)\geq 2^{-n}$$ almost everywhere. To obtain the conclusion for all $$V,s$$, note that the conclusion is determined by a countable dense set of $$V,s$$.

Theorem 16

Let $$E\subset \mathbb R^m$$ satisfy $$\mathcal H^n(E)<\infty$$ and suppose that, for $$\mathcal H^n$$-a.e. $$x\in E$$, $$E$$ has a unique approximate tangent plane at $$x$$. Then $$E$$ is $$n$$-rectifiable.

Proof. By Lemma 11, there exists a decomposition $$E=E'\cup S$$, where $$E'$$ is $$n$$-rectifiable and $$S$$ is purely $$n$$-unrectifiable. We must show that $$\mathcal H^n(S)=0$$. Note that, by applying Lemma 6 to $$E'$$, we see that the approximate tangent plane to $$E$$ at $$x\in S$$ is also an approximate tangent plane to $$S$$ for $$\mathcal H^n$$-a.e. $$x$$.

It suffices to show, for a fixed $$W\in G(n,m)$$, that the set $$S'$$ of $$x\in S$$ whose approximate tangent plane $$V_x$$ lies in $$B(W,\delta)$$ has measure zero. Suppose not. Then, for any $$\lambda>0$$, there exists an $$R>0$$ such that the set $$S''$$ of those $$x\in S'$$ with

$\sup_{0<r<R} \frac{\mathcal H^n(S\cap B(a,r) \setminus C(a,V_a,1/3))}{r^n} < \lambda 3^{-n}$

has positive measure. Fix an $$x \in S''$$. Since $$\|\pi_{V_x}-\pi_{W}\| \leq 1/3$$, for every $$0<r<R$$ we have

$C(x,W^\perp,1/3)\cap B(x,r) \subset B(x,r)\setminus C(x,V_x,1/3).$

Thus, for $$x\in S''$$ and $$0<r<R$$,

$\mathcal H^n(S'\cap C(x,W^\perp,1/3)\cap B(x,r)) < \lambda 3^{-n} r^n.$

If $$\lambda < 240^{-m-1}$$, Corollary 71 implies $$\mathcal H^n(S'')=0$$, a contradiction.

For the remainder of the course, we will prove the Besicovitch projection theorem.

Theorem 17 (Besicovitch projection theorem)

Let $$S\subset \mathbb R^2$$ be purely 1-unrectifiable with $$\mathcal H^1(S)<\infty$$. Then for $$\mathcal H^1$$-a.e. $$e \in \mathbb S^1$$,

$\mathcal L^1(\pi_e(S))=0.$

From Corollary 3, we see that a purely unrectifiable set has many radiating out of almost every point in all directions at almost every point. We now precisely describe two ways in which this can occur.

Definition 25

Let $$S\subset \mathbb R^2$$ and $$x\in S$$. For $$e\in \mathbb S^1$$ let $$l_e(x)$$ be the half line $$x+[0,\infty)e$$ and for $$I\subset \mathbb S^1$$, let $$C(I,x)$$ be the cone $$\bigcup_{e\in I}l_e(x)$$. For $$r>0$$ let $$H_x(r)$$ be those $$e\in \mathbb S^1$$ for which

$|K \cap l_e(x)\cap B(x,r)| \geq 2.$

That is, $$S\cap l_e(x)\cap B(x,r)$$ contains another point of $$K$$. Also let $$H_x=\bigcap_{r>0} H_x(r)$$, the directions that contain other points of $$S$$ arbitrarily close to $$x$$. For $$e\in\mathbb S^1$$, we let $$H_e$$ be those $$x\in S$$ for which $$e\in H_x$$.

For $$R,M,\epsilon>0$$ let $$D_x(R,M,\epsilon)$$ be those $$e\in \mathbb S^1$$ for which there exists $$0<r<R$$ and an interval $$I\subset \mathbb S^1$$ with $$e\in I$$ and $$0<\mathcal H^1(I)<\epsilon$$ such that

$\frac{\mathcal H^1(S\cap C(x,I) \cap B(x,r))}{r} \geq M\mathcal H^1(I).$

That is, the density of $$S$$ in the cone $$C(x,I)$$ at scale $$r$$ is very high, compared to the length of $$I$$. Also let $$D_x = \bigcap_{R,M,\epsilon>0} D_x(R,M,\epsilon)$$. For $$e\in\mathbb S^1$$, we let $$D_e$$ be those $$x\in S$$ for which $$e\in D_x$$.

The main step in proving Theorem 17 is the following.

Proposition 2

Let $$S\subset \mathbb R^2$$ be purely 1-unrectifiable with $$\mathcal H^1(S)<\infty$$. For $$\mathcal H^1$$-a.e. $$x\in S$$, $$\mathcal H^1(\mathbb S\setminus H_x\cup D_x)=0$$.

Before proving Proposition 2, we will demonstrate how it is used to prove Theorem 17.

Lemma 14 (Special case of the coarea formula)

For any $$e\in \mathbb S^1$$ and any compact $$K\subset \mathbb R^2$$,

$\int_{\mathbb R}\operatorname{card}(K\cap l_e(t)) \, \mathrm{d}t \leq \mathcal H^1(K).$

In particular, if $$\mathcal H^1(K)<\infty$$ then for any $$e\in \mathbb S^1$$, $$\mathcal L^1(\pi_{e^\perp}(H_e))=0$$.

Proof. Since $$K$$ is compact,

$f(t) = \operatorname{card}(K\cap l_e(t))$

is a Borel function. Indeed, if, for $$\delta>0$$,

$f_\delta(t) = \max\{n \in \mathbb N: \exists x_1,\ldots,x_n \in K \cap l_e(t) \text{ with } \|x_i-x_j\|\geq \delta\ \forall 1\leq i\neq j\leq n\}$

then $$f_\delta$$ monotonically increases to $$f$$ as $$\delta\to 0$$. Since $$K$$ is compact, the $$f_\delta$$ are lower semi-continuous. Therefore, by the monotone convergence theorem, it suffices to bound the integral of each $$f_\delta$$.

Fix $$\delta>0$$ and cover $$K$$ by sets $$E_1,E_2,\ldots$$ with $$\operatorname{diam}E_i <\delta$$ such that

$\sum_{i\in\mathbb N} \operatorname{diam}E_i \leq \mathcal H^1(K) +\delta.$

Note that

$f_\delta(t) \leq \operatorname{card}(\{i: E_i\cap l_e(t) \neq\emptyset\}).$

Therefore

$\begin{split}\int_\mathbb Rf_\delta \, \mathrm{d}\mathcal L^1 &\leq \int_\mathbb R\sum_{i\in\mathbb N} \chi_{\{(i,t):E_i\cap l_e(t) \neq\emptyset\}}\\ &= \sum_{i\in\mathbb N} \int_\mathbb R\chi_{\{(i,t):E_i\cap l_e(t) \neq\emptyset\}}\\ &\leq \sum_{i\in\mathbb N} \operatorname{diam}E_i\\ &\leq \mathcal H^1(K) +\delta,\end{split}$

as required.

Lemma 15

Let $$S\subset \mathbb R^2$$ be $$\mathcal H^1$$-measurable with $$\mathcal H^1(S)<\infty$$. Then for any $$e\in \mathbb S^1$$, $$\mathcal L^1(\pi_{e^\perp}(D_e))=0$$.*

Proof. Fix $$e\in\mathbb S^1$$. For any $$M\in \mathbb N$$ and $$t \in \pi_{e^\perp}(D_e)$$ there exists an $$x\in D_e$$, $$r_x>0$$ and an interval $$e\in I_x\subset \mathbb S^1$$ with $$\operatorname{diam}I < 1/10$$ such that

$\mathcal H^1(S\cap C(x,I)\cap B(x,r)) \geq M \mathcal H^1(rI).$

Apply Lemma 4 to the intervals $$J_x=\pi_{e^\perp}(C(x,I_x)\cap B(x,r_x))$$ to obtain a disjoint collection $$J_{x_1},J_{x_2},\ldots \subset \mathbb R$$ such that

$D_e \subset \bigcup_{i\in\mathbb N} 5J_{x_i}.$

Therefore

$\begin{split}\mathcal L^1(D_e) &\leq \sum_{i\in\mathbb N} 5\mathcal L^1(J_{x_i})\\ &\leq 5\sum_{i\in \mathbb N} \mathcal H^1(rI_{x_i})\\ &\leq \frac{5}{M} \sum_{i\in \mathbb N} \mathcal H^1(S\cap C(x_i,I_{x_i}) \cap B(x_i,r_{x_i}))\\ &\leq \frac{5}{M} \mathcal H^1(S),\end{split}$

where the final inequality follows from the disjointness of the sets $$C(x_i,I_{x_i})\cap B(x_i,r_i)$$, $$i\in\mathbb N$$. Since this is true for all $$M\in\mathbb N$$, $$\mathcal L^1(D_e)=0$$.

Proof of Theorem 17 using Proposition 2.

Proof. By the inner regularity of $$\mathcal H^1$$, it suffices to prove the result for compact $$S$$. By definition, we have

$\{(x,e)\in S\times \mathbb S^1: e\not\in H_x\cup D_x\}=\{(x,e): x\not\in H_e \cup D_e\}.$

Proposition 2 implies that the left hand expression has $$\mathcal H^1\times \mathcal H^1$$-measure zero and so Fubini’s theorem implies that, for $$\mathcal H^1$$-a.e. $$e\in\mathbb S^1$$, $$\mathcal H^1(S\setminus H_e\cup D_e)=0$$. Therefore, by Lemma 15 and Lemma 14, $$\pi_{e^\perp}(S)=0$$ for $$\mathcal H^1$$-a.e. $$e\in \mathbb S^1$$.

Proof of Proposition 2.

Proof. Fix $$R,M,\epsilon>0$$ and $$x\in S$$ which satisfies the conclusion of Corollary 3. That is,

(26)#$\Theta^{*,1}(S\cap C(x,I),x) \geq c_0 \mathcal H^1(I)$

for every interval $$I\subset \mathbb S^1$$. It suffices to show that $$\mathcal H^1(\mathbb S^1\setminus H_x(R)\cup D_x(R,M,\epsilon)) =0$$. In fact, we will show that, for any $$e\in \mathbb S^1$$,

$\Theta^{*,1}(H_x(R),e)>0 \quad \text{or} \quad \Theta_*^1(D_x(R,M,\epsilon),e)>0,$

from which the result follows by the Lebesgue density theorem.

To this end, fix $$e\in \mathbb S^1$$ with $$\Theta^{*,1}(H_x(R),e)=0$$. Then for all sufficiently small intervals $$I$$ with $$e\in I \subset \mathbb S^1$$,

(27)#$\mathcal H^1(H_x(R)\cap I) < c_0 \mathcal H^1(I)/4M.$

Fix such an $$I$$. By (26), there exists $$r<R$$ with

(28)#$\mathcal H^1(S\cap C(x,I)\cap B(x,r)) \geq c_0 r \mathcal H^1(I).$

Note that (27) implies

(29)#$\mathcal H^1(H_x(r)\cap I) < c_0\mathcal H^1(I)/4M.$

We will prove that

(30)#$\mathcal H^1(D_x(R,M,\epsilon)\cap I) \geq c_0\mathcal H^1(I)/4M.$

Since $$I$$ is any sufficiently small interval containing $$e$$, this implies

$\Theta^{1}_*(D_x(R,M,\epsilon),e)\geq c_0/4M>0$

as required.

By (29) we may cover $$H_x(r)\cap I$$ by disjoint intervals $$I_1,I_2,\ldots$$ with

$\sum_{i\in \mathbb N} \mathcal H^1(I_i)< c_0\mathcal H^1(I)/4M$

(indeed, the disjointness of the intervals is shown in Example 53). By the definition of $$H_x(r)$$, we know that

(31)#$S\cap C(x,I) \cap B(x,r) \subset \bigcup_{i\in\mathbb N} S\cap C(x,I_i)\cap B(x,r).$

Let $$\mathcal G$$ be those $$i\in\mathbb N$$ with

(32)#$\frac{\mathcal H^1(S\cap C(x,I_i)\cap B(x,r))}{r} \geq M\mathcal H^1(I_i).$

Note that by (28) and (31),

$\begin{split}\sum_{i\in \mathcal G} \frac{\mathcal H^1(S\cap C(x,I_i)\cap B(x,r))}{r} &\geq \frac{\mathcal H^1(S\cap C(x,I)\cap B(x,r))}{r} - \frac{c_0 \mathcal H^1(I)}{4}\\ &\geq \frac{3c_0 \mathcal H^1(I)}{4}.\end{split}$

That is, the cones associated to $$G:=\bigcup_{i\in\mathcal G}I_i$$ cover a large proportion of $$S\cap C(x,I) \cap B(x,r)$$. Moreover, $$G\subset D_x(R,M,\epsilon)$$, because if $$\xi\in G$$ then $$\xi \in I_i$$ for some $$i\in \mathcal G$$ which satisfies the definition of $$D_x(R,M,\epsilon)$$.

Thus, to show (30), it would be enough to bound $$\mathcal H^1(G)$$ from below by a multiple of $$\mathcal H^1(I)$$. But this is not necessarily true: the intervals that form $$G$$ could be extremely thin compared to $$I$$. To accommodate this, we enlarge the intervals $$I_i$$ with $$i\in\mathcal G$$ as follows. For each $$i\in \mathcal G$$ enlarge $$I_i$$ until (32) becomes an equality or until $$I_i$$ intersects another $$I_j$$. If the first possibility occurs then we still have $$I_i\subset D_x(R,M,\epsilon)$$. If the second possibility occurs then we merge the two intervals; since both sides of (32) are linear in $$I$$, and the boundary of each $$C(x,I_i)$$ contains no points of $$S$$, (32) remains true after the merge. By the same reasoning, both sides of (32) are continuous under expanding $$I$$ and consequently one of the two possibilities must occur.

This results in a disjoint collection of intervals $$\tilde I_i$$ for which (32) is an equality. Moreover,

$G= \bigcup_{i\in \mathcal G}I_i \subset \bigcup \tilde I_i$

and, by construction, each $$\tilde I_i \in D_x(R, M, \epsilon)$$. Therefore

$\begin{split}\mathcal H^1(D_x(r_0,\epsilon,M) \cap I) &\geq \sum \mathcal H^1(\tilde I_i)\\ &= \sum \frac{\mathcal H^1(S\cap B(x,r) \cap C(x,\tilde I_i))}{rM}\\ &\geq \sum_{i\in\mathcal G}\frac{\mathcal H^1(S\cap B(x,r) \cap C(x,I_i))}{rM}\\ &\geq \frac{3c_0\mathcal H^1(I)}{4M}.\end{split}$

## Exercises#

Example 48

Prove that the coordinate projections of the four corner Cantor set $$K\subset \mathbb R^2$$ have Lebesgue measure zero.

Example 49

A second proof that the four corner Cantor set is purely 1 unrectifiable.

Suppose that $$f\colon \mathbb R\to \mathbb R^2$$ satisfies $$\mathcal H^1(f(\mathbb R)\cap K)>0$$.

1. Prove that there exists $$x\in \mathbb R$$ that is a density point of $$f^{-1}(K)$$ such that $$f'(x)\neq 0$$.

2. Therefore, for sufficiently small $$r$$, $$f(x-r,x+r)$$ is approximated by a line segment of length $$2rf'(x)$$ that is mostly contained in $$K$$. Derive a contradiction.

Example 50

Prove that the decomposition given in Lemma 11 is unique up to $$\mathcal H^n$$-null sets.

Example 51

Think about Corollary 3 and Proposition 2 in regard to the four corner Cantor set.

Example 52

Let $$K\subset \mathbb R^2$$ be compact. Show that

$\{(x,e): e \in H_x\} \quad \text{and} \quad \{(x,e): e \in D_x\}$

are Borel subsets of $$K\times \mathbb S^1$$.

Example 53

Show that in the definitions of $$\mathcal L^n$$ and $$\mathcal H^n$$ we may suppose the covering intervals, respectively sets, may be chosen to be disjoint.

Example 54 (Open problem)

For $$k\geq 2$$, does there exist a compact purely 1-unrectifiable $$S\subset \mathbb R^2$$ with $$\mathcal H^1(S)>0$$ that intersects every line in at most $$k$$ points?